SOLUTION: Here's my question.
"An aircraft travels at a speed of 550mph during the 1st half of it's journey and 500mph during the second half of it's journey. Total flight time is 05:00 h
Algebra.Com
Question 455551: Here's my question.
"An aircraft travels at a speed of 550mph during the 1st half of it's journey and 500mph during the second half of it's journey. Total flight time is 05:00 hours. Based on this determine the distance between the airport of departure and arrival."
Sounds straightforward enough, my answer was 2625 miles. However the correct answer is 2619 miles?! I need clarification here, am I right or do I need to loose 6 miles somehow?
Pete M
Found 2 solutions by ankor@dixie-net.com, richwmiller:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
An aircraft travels at a speed of 550mph during the 1st half of it's journey and
500mph during the second half of it's journey.
Total flight time is 05:00 hours.
Based on this determine the distance between the airport of departure and arrival."
:
Let d = distance between the airports
:
Write a time equation: time = dist/speed
:
550mph time + 500mph time = 5 hrs
+ = 5
multiply by 5500,
5500* + 5500 = 5500(5)
cancel the denominators
10(.5d) + 11(.5d) = 27500
5d + 5.5d = 27500
10.5d = 27500
d =
d = 2619 mi
:
:
You can prove this to yourself, find the time at each speed
1309.5/550 = 2.38 hrs
1309.5/500 = 2.62 hrs
----------------------
total time: 5.0 hrs
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
You used half of the time to arrive at 2625.
500*2.5+550*2.5=2625
The misleading part is half of the journey.
Is that half by time or distance?
550*x=d
500*y=d
since the distances are the same
550x=500y
x+y=5
x = 50/21, y = 55/21, z = 55000/21
x=2.38095, y=2.61905, z=2619.05
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